EIE209 Basic Electronics
First-order transient
Contents
• Inductor and capacitor
• Simple RC and RL circuits
• Transient solutions
Prof. C.K. Tse: First Order Transient
Constitutive relation
®
An electrical element is defined by its relationship between v and i.
This is called constitutive relation. In general, we write
v = f (i )
or
i = g (v )
i
+ v –
®
®
For a resistor,
® v = i R
The constitutive relation of a resistor has no dependence upon time.
Prof. C.K. Tse: First Order Transient
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Capacitor and inductor
ic
C
The constitutive relation of a
linear capacitor is:
+ vc –
iL
dv
ic = C c
dt
where the proportionality
constant C is capacitance
(unit is farad or F)
The constitutive relation of a
linear inductor is:
L
+ vL –
vL = L
di L
dt
Prof. C.K. Tse: First Order Transient
where the proportionality
constant L is inductance
(unit is henry or H)
3
What happens if a circuit has C and/or L?
®
The circuit becomes dynamic. That means:
®
®
®
Its behaviour is a function of time.
Its behaviour is described by a (set of) differential
equation(s).
It has a transient response as well as a steady state.
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Resistive circuits have no transient
®
®
Consider a resistive circuit.
When the switch is turned on,
the voltage across R becomes V
immediately (in zero time).
®
v=V=iR
for all t > 0
®
i =V/R
for all t > 0
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A simple first-order RC circuit
®Let
us consider a very simple dynamic circuit, which
contains one capacitor.
®After
t = 0, the circuit is closed. So, we can easily write
®and
®Thus,
we have
®Thus,
we have
®If
Vo for t>0
fi
the initial condition is vC(0+) = 0, then A = –Vo.
®Thus,
the solution is
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Transient response of the RC circuit
®Once
we have the capacitor voltage, we can find
anything.
®Starting
®We
with
can derive the current as
time constant
®We
see the solution typically has a TRANSIENT
which dies out eventually, and as t tends to ∞, the
solution settles to a steady state.
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A simple first-order RL circuit
®Consider
a RL circuit.
®Before t = 0, the switch is closed (turned on).
Current goes through the switch and nothing goes
to R and L. Initially, iL(0–) = 0.
®At t = 0, the switch is opened. Current goes to R
and L.
®We know from KCL that Io = iR + iL for t > 0, i.e.,
®The
constitutive relations give
®Hence,
®
fi
®The
From the initial condition, we
have iL(0–) = 0. Continuity of
the inductor current means that
iL(0+) = iL(0–) = 0. Hence,
A = – Io
Thus,
solution is
Prof. C.K. Tse: First Order Transient
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Transient response of the RL circuit
®
Starting with
®
We can find vL(t):
time constant
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Observation — first-order transients
®
First order transients are always like these:
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Let’s do some math
x( t )
x( t )
x(t) = 5(1 –
x( t )
e–t/t)
6
5
t
0
x(t) = 1 + 5(1 – e–t/t)
5
t
–2
x( t )
x(t) = 5 e–t/t
x( t )
6
5
x(t) = 1 + 5 e–t/t
4
0
t
0
1
0
x( t )
x(t) = –2 + 7(1 – e–t/t)
t
1
0
x(t) = –3 + 7 e–t/t
t
t
0
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General first-order solution
NO NEED TO SOLVE ANY EQUATION, just find
1.
2.
3.
the starting point of capacitor voltage or inductor current
the ending point of ………… ………. ……. ………. ……….
the time constant t
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Finding t
For the simple first-order RC circuit,
For the simple first-order RL circuit,
t = C R.
t = L / R.
The problem is
Given a first-order circuit (which may look complicated),
how to find the equivalent simple RC or RL circuit.
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A quick way to find t
Since the time constant is independent of the sources, we first of all set
all sources to zero. That means, short-circuit all voltage sources and opencircuit all current sources. Then, reduce the circuit to
either
Req
Ceq
or
Req
Leq
Example:
R1
R1
t = C (R1 || R2)
+
–
R2
C
R2
C
R1 || R2
Prof. C.K. Tse: First Order Transient
C
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Example 1 (boundary conditions given)
Find vc(t) for t > 0 without solving any
differential equation.
Step 1: initial point (given)
vc(0–) = 50 V is known (but not what we want).
Continuity of cap voltage guarantees that
vc(0+) = vc(0–) = 50 V.
Step 2: final point (almost given)
vc(t) = –20 + 70 e–t/CR
vc(∞) = –20 V.
Step 3: time constant
The equivalent RC circuit is:
Thus, t = CR.
Answer is:
50
0
–20
Prof. C.K. Tse: First Order Transient
t
15
Example 2 (non-trivial boundary conditions)
t=0
Find v1(t) and v2(t) for t > 0 without solving
any differential equation.
v1(0–)
Suppose
= 5 V and v2
= 2 V.
Problem: how to find the final voltage values.
(0–)
+
v1
–
i1
C1
2F
R = 1Ω
C2
3F
i2 +
v2
–
Form 7 solution:
You considered the charge transfer Dq from C1 to C2.
Dq
++
++++++
q1 fi q1 – Dq
++++
q2 fi q2 + Dq
Prof. C.K. Tse: First Order Transient
Use charge balance
and KVL equations
to find the final
voltage values.
Clumsy solution!
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Example 2 (elegant solution)
t=0
We need not consider CHARGE!
Step 1: initial point (given)
v1(0–) = 5 V and v2(0–) = 2 V are known.
Continuity of cap voltage guarantees that
v1(0+) = 5 V and v2(0+) = 2 V.
Step 2: final point (non-trivial)
C1:
C2:
dv 1
dt
dv
i2 = 3 2
dt
i1 = 2
for all t
for all t
dv
dv
After t>0, we have i1 = –i2, i.e., 2 1 + 3 2 = 0
dt
dt
fi 2v1(t) + 3v2(t) = K
for all t > 0.
Prof.
Integration constant
+
v1
–
i1
C1
2F
R = 1Ω
C2
3F
i2 +
v2
–
At t = 0+, this equation means
2*5 + 3*2 = K. Thus, K = 16.
Thus,
2v1(t) + 3v2(t) = 16 for t > 0.
At t =∞, we have v1(∞)=v2(∞)
from KVL. Hence,
2v1(∞)+3v1(∞)=16
fi v1(∞)=v2(∞)=16/5 V.
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Example 2 (elegant solution)
R = 1Ω
Step 3: time constant
The circuit after t = 0 is
This can be reduced to
+
v1
–
i1
C1
2F
C1 C2
C1 + C 2
The time constant is
t=
C2
3F
= 6/5 F
i2 +
v2
–
R = 1Ω
C1C 2
6
¥R =
s
C1 + C 2
5
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Example 2 (answer)
v1
v2
5V
16/5=3.2V
16/5=3.2V
2V
t
t
v 2 (t ) = 2 + 1.2(1 - e -5t / 6 ) V
v1 (t ) = 3.2 + 1.8e -5t / 6 V
We can also find the current by
i(t ) =
v 1(t ) - v 2 (t )
= 3e -5t / 6 A
R
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General procedure
®
Set up the differential equation(s) for the circuit in terms of capacitor
voltage(s) or inductor current(s).
®
The rest is just Form 7 Applied Math!
2
d vc
dv c
+ Bv c = C
dt
In the previous example:
®
E.g.,
®
Get the general solution.
®
There should be n arbitrary constants for an nth-order circuit.
®
Using initial conditions, find all the arbitrary constants.
d t2
+A
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Basic question 1
®Why
must we choose capacitor voltage and inductor current as the
variable(s) for setting up differential equations?
®
Never try to set differential equation in terms of other kinds of variables!
dv R
+ kv R = Vo
dt
®Answer:
®Capacitor
voltages and inductor currents are guaranteed to be CONTINUOUS
before and after the switching. So, it is always true that
vC (0- ) = v C (0 + )
and
iL (0 - ) = iL (0+ )
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Basic question 1
®Then,
why capacitor voltages and inductor currents are guaranteed to be
continuous?
®Answer:
try to prove it by contradiction. Suppose vc and iL are discontinuous at t =
0. That means,
vC (0- ) ≠ v C (0 + ) and iL (0 - ) ≠ iL (0+ )
vc or iL
®Let’s
®Now,
recall the constitutive relations.
di
dv
and
vL = L L
ic = C c
dt
dt
®Then, we have
iC Æ • and v L Æ •
slope = ∞
t
which is not permitted in the physical world.
®So,
capacitor voltages and inductor currents must not be discontinuous.
Prof. C.K. Tse: First Order Transient
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Basic question 2
®How
to get the differential equation systematically for any circuit?
®For
simple circuits (like the simple RC and RL circuits), we can get it by an ad
hoc procedure, as in the previous examples. But, if the circuit is big, it seems
rather difficult!
®Hint:
®Graph
theory. (See Chapter 8 of my book)
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